Superflous Symbols

Relevant wiki: System of Equations - Substitution

Adding up gives $(x+y+z)^2=26+27+28=81$ , so $x+y+z=\pm 9$ and $(x,y,z)=\pm\left(\frac{26}{9},\frac{27}{9},\frac{28}{9}\right)$ . There are $\boxed{2}$ solutions.

$x(x+y+z)=26\implies$ Eq.(1)

$y(x+y+z)=27\implies$ Eq.(2)

$z(x+y+z)=28\implies$ Eq.(3)

Notice that for these equations, $x,y,z$ and $(x+y+z)$ all cannot be $0$ . Therefore, it is safe to divide these equations:

Eq.(2) $\div$ Eq.(1):

$\dfrac{y(x+y+z)}{x(x+y+z)} = \dfrac{27}{26}\\ \dfrac{y}{x} = \dfrac{27}{26}$

$y=\dfrac{27}{26}x\implies$ Eq.(4)

Eq.(3) $\div$ Eq.(1):

$\dfrac{z(x+y+z)}{x(x+y+z)} = \dfrac{28}{26}\\ \dfrac{z}{x} = \dfrac{28}{26}$

$z=\dfrac{28}{26}x\implies$ Eq.(5)

Substitute Eq.(4) and Eq.(5) into Eq.(1):

$x\left(x+\dfrac{27}{26}x + \dfrac{28}{26}x \right)=26\\ x\left(\dfrac{26+27+28}{26}x\right) = 26\\ \dfrac{81}{26}x^2=26\\ x^2 = \dfrac{26^2}{81}\\ x=\pm \sqrt{\dfrac{26^2}{81}} = \pm \dfrac{26}{9}$

When $x=\dfrac{26}{9}$

$y=\dfrac{27}{26} \left(\dfrac{26}{9}\right) = \dfrac{27}{9} = 3 \quad\quad\quad z=\dfrac{28}{26} \left(\dfrac{26}{9}\right) = \dfrac{28}{9}$

When $x=-\dfrac{26}{9}$

$y=\dfrac{27}{26} \left(-\dfrac{26}{9}\right) = -\dfrac{27}{9} = -3 \quad\quad\quad z=\dfrac{28}{26} \left(-\dfrac{26}{9}\right) = -\dfrac{28}{9}$

The solutions are: $(x,y,z)=\left(\dfrac{26}{9},3,\dfrac{28}{9}\right),\;\left(-\dfrac{26}{9},-3,-\dfrac{28}{9}\right)$

Number of triples that satisfies the equations $= \boxed{2}$